Desafío: libreta de calificaciones

Comment pouvons-nous vous aider ?

Fernando Casabuena

4 mars 2017 14:41

Hola, estoy siguiendo el curso de SQL pero me he quedado bloqueado en ésta parte, el enunciado es:

Creamos una base de datos para llevar el seguimiento de las calificaciones de los estudiantes, con su nombre, calificación numérica y qué porcentaje de actividades han completado. En este primer paso, selecciona todos los renglones y despliega name, number_grade y percent_completed, el cual puedes calcular multiplicando y redondeando la columna fraction_completed.

Supongo que para calcular percent_completed hay que hacer la multiplicación number_grade * fraction_completed

Pero no sé como hacer esto.

Alguien me puede orientar?

Saludos

Hello, I am following the course of SQL but I have been blocked in this part, the statement is:

We created a database to keep track of students' grades, with their name, numerical rating and what percentage of activities they completed. In this first step, select all rows and display name, number_grade and percent_completed, which you can calculate by multiplying and rounding the fraction_completed column.

I guess to calculate percent_completed you have to do the multiplication number_grade * fraction_completed

But I do not know how to do this.

Can someone guide me?





regards

Nouvelle publication

2 commentaires

Cristian Martin Sguazzin

15 juillet 2017 14:28
0

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Hi man como estas?

Bueno la solucion es esta :

CREATE TABLE student_grades (
id INTEGER PRIMARY KEY AUTOINCREMENT,
name TEXT,
number_grade INTEGER,
fraction_completed REAL
);

INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Winston", 90, 0.805);
INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Winnefer", 95, 0.901);
INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Winsteen", 85, 0.906);
INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Wincifer", 66, 0.7054);
INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Winster", 76, 0.5013);
INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Winstonia", 82, 0.9045);

/* PORCENTAJE PASO 1 */

SELECT name, number_grade, round((fraction_completed*100)) as percent_completed
from student_grades;

Saludos !
Cristian Martin Sguazzin

15 juillet 2017 15:14
0

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El segundo paso del ejercicio te lo dejo tambien :

CREATE TABLE student_grades (
id INTEGER PRIMARY KEY AUTOINCREMENT,
name TEXT,
number_grade INTEGER,
fraction_completed REAL
);

INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Winston", 90, 0.805);
INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Winnefer", 95, 0.901);
INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Winsteen", 85, 0.906);
INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Wincifer", 66, 0.7054);
INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Winster", 76, 0.5013);
INSERT INTO student_grades (name, number_grade, fraction_completed)
VALUES ("Winstonia", 82, 0.9045);

SELECT name, number_grade, round((fraction_completed*100)) as percent_completed
from student_grades;

SELECT count(*), /*LINEA COMENTADA -->*/ /*number_grade,*/

CASE
WHEN number_grade > 90 THEN "A"
WHEN number_grade > 80 THEN "B"
WHEN number_grade > 70 THEN "C"
ELSE "F"
END as "letter_grade"
FROM student_grades
group by letter_grade;

Saludos !

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Desafío: libreta de calificaciones

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2 commentaires

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